Math, asked by whitepearl434, 7 months ago

solve the following limit​

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Answers

Answered by vsksen12
1

Answer:

final answer is 6√5

for step cheak attachment

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Answered by Anonymous
4

Answer:

6√5

Step-by-step explanation:

By directly substituting the limits x = 3, we get (0/0) which is an indeterminate quantity.

Let's solve it by using the method of rationalisation.

Rationalising factor for the denominator of given function is \sqrt{3x-4} + \sqrt{x+2}, so multiply both numerator and denominator by this factor.

{ \implies  \lim \limits_{x \to 3}  \dfrac{ {x}^{2}  - 9}{ \sqrt{3x - 4} -  \sqrt{x + 2}  }  \times  \dfrac{ \sqrt{3x - 4}  +  \sqrt{x + 2} }{ \sqrt{3x - 4 +  \sqrt{x + 2} } } }

{ \implies  \lim \limits_{x \to 3}  \dfrac{(x + 3)(x - 3)(\sqrt{3x - 4}  +  \sqrt{x + 2})}{ (\sqrt{3x - 4})^{2}  - ( \sqrt{x + 2}  )^{2} } }

{ \implies  \lim \limits_{x \to 3}  \dfrac{(x + 3)(x - 3)( \sqrt{ 3x - 4}  +  \sqrt{x + 2})}{3x - 4  - x - 2 } }

{ \implies  \lim \limits_{x \to 3}  \dfrac{(x + 3)(x - 3)( \sqrt{ 3x - 4}  +  \sqrt{x + 2})}{2x - 6 } }

{ \implies  \lim \limits_{x \to 3}  \dfrac{(x + 3) \cancel{(x - 3)}( \sqrt{ 3x - 4}  +  \sqrt{x + 2})}{2 \cancel{(x - 3)}} }

{ \implies  \lim \limits_{x \to 3}  \dfrac{(x + 3)( \sqrt{ 3x - 4}  +  \sqrt{x + 2})}{2} }

{ \implies \dfrac{(3+ 3)( \sqrt{ 3(3) - 4}  +  \sqrt{3 + 2})}{2} }

{ \implies \dfrac{(6)( \sqrt{9 - 4}  +  \sqrt{5})}{2} }

{ \implies \dfrac{6( \sqrt{5}  +  \sqrt{5})}{2} }

{ \implies \dfrac{6( 2\sqrt{5})}{2} }

 \implies6 \sqrt{5}

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